Using the microscopic form of Ohm's law `(J=sigma E)` , prove that conductivity of a metal can be written as `sigma="ne"mu`. Here n is the number of free electrons per unit volume and `mu` is mobility of free electrons . In case of semiconductors there are two types of conduction particles, one is free electrons and the other is known as a hole. Charge on the hole may be assumed to be equal and opposite of that on electron. The number density of free electrons and holes in the semiconducting material are n and p respectively. Assuming `mu_e and mu_h` as mobility of free electrons and holes respectively , write the exopression of conductivity of the semiconducting material.

Explanation

We know that current density is given by
`J=sigmaE`
Also mobility `mu=(v_d)/E rArr E=(v_d)/(mu)`
Therefore `J=sigma(v_d)/(mu)`
As `I="neA"v_d`
`therefore v_d=I/("neA") rArr J="ne"v_d (because J=IA)`
`rArr "ne" mu =sigma` ......(i)
For the semiconducting material,
`mu_e`= mobility of free electrons
`mu_h`= mobility of free holes
n= number density of electrons
p= number density of holes
`sigma=` conductivity of material
`therefore` Conductivity `sigma=e(nmu_e+pmu_h)`
(Using (i))